[QECGatesCost] strict & account for more leaf bloqs #1313
Conversation
|
I don't see a comment, did you try to leave one? From offline: I'll use machine precision to compare the angles |
| if isinstance(b, (Rz, Rx, Ry)): | ||
| if is_symbolic(b.angle): | ||
| return False # Symbolic rotation | ||
| if np.any(np.abs(b.angle - _T_ANGLES) < _ANGLE_ATOL): | ||
| return True # T hidden in a rotation bloq | ||
| return False |
There was a problem hiding this comment.
I think bloq_is_t_gate should return True only if the unitary of the bloq is equal to that of a T gate. For other single qubit rotations, we should have default decompositions; which a user can choose to override if needed; to avoid undercounting cliffords.
There was a problem hiding this comment.
we generally don't have "default" decompositions that one can override.
I can change this function name to bloq_is_t_like. Counting Rx(pi/4) as 1 T gate is much more accurate than 1 "rotation", so I'm going to proceed with the change. We can add a decomposition of Rx to Rz later or change the workflow where a re-write rule is responsible for simplifying any Rx(theta) to X/4 or whatever we like (and remove this itchy logic from the gate counting code).
There was a problem hiding this comment.
it's also unclear (to me) how we'd want to count cliffords for a situation like Rx(pi/4). Using an auto-correcting circuit like Fig 17 of game-of-surface-codes implies you can do any pauli with the same number of gates
The first commit is #1311 ; which can't be merged independently because it reveals errors that this PR fixes.